C ++条件编译 [英] C++ conditional compilation

问题描述

我有以下代码片段：

#ifdef DO_LOG
#define log(p) record(p)
#else
#define log(p)
#endif

void record(char *data){
.....
.....
}

现在如果我调用 log（hello world）在我的代码和 DO_LOG 没有定义，行会被编译，换句话说， up字符串hello world的内存？

Now if I call log("hello world") in my code and DO_LOG isn't defined, will the line be compiled, in other words will it eat up the memory for the string "hello world"?

在程序中有很多记录调用，它是内存敏感的，所以有什么其他方法有条件地编译，以便它只依赖于 #define DO_LOG ？

P.S. There are a lot of record calls in the program and it is memory sensitive, so is there any other way to conditionally compile so that it only depends on the #define DO_LOG?

推荐答案

这应该是微不足道的，通过检查生成的二进制文件来验证。

This should be trivial to verify for yourself by inspecting the resulting binary.

我会说no，因为表达式完全消失了，编译器将永远不会看到字符串（它被预处理器的宏扩展删除）。

I would say "no", since the expression totally goes away, the compiler will never see the string (it's removed by the preprocessor's macro expansion).

这篇关于C ++条件编译的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！